If the graph of 27 x + 33 y = 297 is shifted down 5 units

Choice A is correct. When the graph of an equation in the form $Ax+By=C$, where $A$, $B$, and $C$ are constants, is shifted down $k$ units in the xy-plane, the resulting graph can be represented by the equation $Ax+B\left(y+k\right)=C$. It's given that the graph of $27x+33y=297$ is shifted down $5$ units in the xy-plane. Therefore, the resulting graph can be represented by the equation $27x+33\left(y+5\right)=297$, or $27x+33y+165=297$. Subtracting $165$ from both sides of this equation yields $27x+33y=132$. The y-intercept of the graph of an equation in the xy-plane is the point where the line intersects the y-axis, represented by the point $\left(0,y\right)$. Substituting $0$ for $x$ in the equation $27x+33y=132$ yields $27\left(0\right)+33y=132$, or $33y=132$. Dividing both sides of this equation by $33$ yields $y=4$. Therefore, if the graph of $27x+33y=297$ is shifted down $5$ units, the y-intercept of the resulting graph is $\left(0,4\right)$.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the y-intercept of the graph of $27x+33y=297$ shifted up, not down, $5$ units.

Choice D is incorrect and may result from conceptual or calculation errors.